Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The kinetic energy calculation for any meteor would be the traditional one half times the mass times the velocity squared. In terms of the events that occur on the ground, does it matter whether the meteor explodes in the atmosphere or doesn't explode? Is the energy released to the atmosphere and structures and people on the ground by the shock wave it creates the same whether it explodes in the air or not?

share|cite|improve this question
That shockwave, btw, was from the explosion in the air. A small stone hitting the ground at terminal velocity... think of it this way - if you drop a rock (even a large boulder) off the Empire State Building, it wouldn't shatter a single window, much less every window in the city. – Chris White Feb 23 '13 at 18:17

The energy released is roughly the same regardless of the altitude of the explosion, and it is delivered to the atmosphere as well as objects on the ground. For higher altitudes, however, more and more of the energy is taken up by the atmosphere, and less of it reaches the ground.

The Tunguska blast offers a good comparison. Though it was admittedly an order of magnitude stronger (10 to 15 MT) than the 500 kT of the Chelyabinsk blast, it happened much lower down (at 5 to 10 km altitude instead of 15 to 25), and caused far more destruction, felling about 80 million trees.

Another (shaky) comparison is with the Hiroshima nuclear blast, which was weaker (16 kT) and much lower down (600 m). The explosion itself caused the huge material destruction, even if it was the radiation that arguably caused most of the deaths and subsequent horrors.

share|cite|improve this answer

Total Energy is the same, but the location is not, which makes a big difference.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.